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  • Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 15

Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 15

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Answer: \tan \left(\frac{y}{x}\right)=\log \left(\frac{e}{x}\right)

Given: The angle is \left(\frac{y}{x}-\frac{y}{x}\right)

To find: We have to find the equation of the curve which passes through \left(1, \frac{\pi}{4}\right)

Hint: First take the slope of the curve i.e.\frac{d y}{d x}=\tan \theta and take a linear equation y=v x

Solution:

The slope of the curve is \frac{d y}{d x}=\tan \theta

We have,

        =\theta=\left[\frac{y}{x}-\frac{y}{x}\right]

        \begin{aligned} &\text { Then } \begin{array}{l} \frac{d y}{d x}=\text { tan inverse tan }\left\{\left[\frac{y}{x}-\frac{y}{x}\right]\right\} \\\\ \quad=\frac{d y}{d x}=\frac{y}{x}-\frac{y}{x} \end{array} \\ &\text { Let } y=v x \end{aligned}

Differentiating with respect to x we get

        \begin{aligned} &\frac{d y}{d x}=v+x \frac{d v}{d x} \\\\ &x \frac{d v}{d x}=\frac{d y}{d x}-v \end{aligned}

        x \frac{d v}{d x}=\frac{d y}{d x}-\frac{y}{x} \quad\quad\quad\quad\left[v=\frac{y}{x}\right]

        x \frac{d v}{d x}=-v        [From equation i]

Integrating on both sides

        \begin{aligned} &\int v d v=\int-\frac{1}{x} d x \\\\ &\tan \frac{y}{x}=-\log |x|+C \ldots(i i) \end{aligned}

The curve passes through\left[1, \frac{\pi}{4}\right] it satisfices equation (ii)

        \begin{aligned} &=\log \frac{\pi}{4}=-\log |1|+C \\\\ &=C=1 \end{aligned}

\begin{aligned} &\text { Put } C=1, \tan \frac{y}{x}=-\log |x|+1 \\\\ &=\tan \frac{y}{x}=-\log |x|+1 \\\\ &=\tan \frac{y}{x}=\log \left|\frac{e}{x}\right| \end{aligned}

Hence, required equation is found.

 

 

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